Matrices and Determinants
Gauss-Jordan Reduction
Carrying the elimination further so the solution is read directly from the matrix, with no back-substitution at all.
What this page covers
- Distinguish row-echelon from reduced row-echelon form
- Clear above the pivots as well as below
- Read a solution directly from a reduced matrix
- Say when the extra work is and is not worth it
One step further
Gaussian elimination stops at row-echelon form and finishes by back-substitution. Gauss-Jordan continues the row operations until the coefficient block is the identity matrix, at which point the solution is simply written in the augmented column.
| Gaussian | Gauss-Jordan | |
|---|---|---|
| Clears | Below each pivot only | Below and above |
| Ends at | Row-echelon form | Reduced row-echelon form |
| Left block becomes | Upper triangular | The identity matrix |
| Finishing step | Back-substitution | None — just read it off |
| Best for | Solving one system | Inverting; several right-hand sides |
Worked example: the source's 2×2 case
Solve x - 3y = 4, 2x + y = 1
| 1 | -3 | 4 |
| 2 | 1 | 1 |
Subtract twice row 1 from row 2, then divide by 7:
| 1 | -3 | 4 |
| 0 | 1 | -1 |
Gauss-Jordan continues. The source's instruction is to multiply the second row by 3 and add to the first, clearing the -3 above the second pivot:
| 1 | 0 | 1 |
| 0 | 1 | -1 |
No back-substitution was needed. The left block is the identity matrix, so each row reads as a single variable equalling a single number.
Worked example: three variables
The source's 3×3 case
Starting from the row-echelon form obtained earlier for x + y + z = 5, x - 4y - 3z = 11, -2x + 2y + 5z = -30:
| 1 | 1 | 1 | 5 |
| 0 | 1 | 4/5 | -6/5 |
| 0 | 0 | 1 | -4 |
The source works upward, describing it as clean up the rightmost column, then clean up the middle column.
| 1 | 1 | 0 | 9 |
| 0 | 1 | 0 | 2 |
| 0 | 0 | 1 | -4 |
| 1 | 0 | 0 | 7 |
| 0 | 1 | 0 | 2 |
| 0 | 0 | 1 | -4 |
7 + 2 - 4 = 5 ✓; 7 - 8 + 12 = 11 ✓; -14 + 4 - 20 = -30 ✓.
Work upward from the last pivot, clearing one column at a time. Going in the other order would undo work already done.
Row operations as matrix multiplication
The source makes a point in Week 12 that is easy to miss and worth the effort: every row operation can be performed by multiplying on the left by a suitable matrix.
The source's demonstration
Premultiplying the augmented matrix by
| 1 | 0 |
| -2 | 1 |
Premultiplying by
| 1 | 0 |
| 0 | 1/7 |
Premultiplying by
| 1 | 3 |
| 0 | 1 |
The source then multiplies all three together into a single matrix and applies it in one step:
| 1/7 | 3/7 |
| -2/7 | 1/7 |
Applying this to the augmented matrix takes it straight to reduced form. The source labels the observation achieving row operations using matrix multiplication.
That combined matrix is the inverse of the coefficient matrix. This is not a coincidence — it is exactly why the Gauss-Jordan method inverts a matrix, and it is developed in Matrix Inverses.
When to use which
| Task | Method | Reason |
|---|---|---|
| Solve one system by hand | Gaussian | Back-substitution is quicker than clearing upward |
| Invert a matrix | Gauss-Jordan | Reduce [A | I] to [I | A-1] |
| Several right-hand sides | Gauss-Jordan | One reduction serves all of them |
| Determine consistency only | Gaussian | The bottom row settles it |
| Machine computation | Gaussian with pivoting | Fewer operations and better numerical stability |
For hand calculation on a single system the extra clearing is rarely worth it. Its value appears when the same coefficient matrix must be used repeatedly.
Common mistakes
| Mistake | Correct | Why |
|---|---|---|
| Clearing upward before finishing downward | Complete row-echelon form first | Otherwise work is undone |
| Clearing columns left to right when going up | Work from the last pivot backwards | Later columns must be cleared first |
| Not applying an operation to the augmented column | The whole row moves | The constants must travel too |
| Expecting the identity when the system is dependent | A zero row will appear instead | The identity requires a unique solution |
| Using Gauss-Jordan on a single hand system | Gaussian is quicker | Back-substitution costs less |
| Multiplying on the right for a row operation | Row operations are left multiplication | Right multiplication performs column operations |
Frequently asked questions
How does it differ from Gaussian elimination?
Gaussian elimination clears only below each pivot and finishes with back-substitution. Gauss-Jordan clears above as well, leaving the identity matrix on the left and the solution on the right.
Is it faster?
For solving one system, no — the extra clearing costs more than the back-substitution it replaces. It pays off when inverting a matrix or handling several right-hand sides at once.
What does the final matrix look like?
The identity on the left of the rule and the solution values on the right. The source's example ends as [1 0 | 1; 0 1 | -1], read straight off as x = 1, y = -1.
Can row operations be done by matrix multiplication?
Yes. Each operation corresponds to multiplying on the left by an elementary matrix, and the source demonstrates this explicitly in Week 12.
Source. Handwritten teaching notes, Week 11 page 7 and Week 12 pages 7-9.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.